Problem: Rewrite the function by completing the square. $g(x)= x^{2} - x -6$ $g(x)=$
Solution: $\begin{aligned} g(x)&= x^2 - x -6 \\\\ &= \left(x^2 - x\right) -6 \end{aligned}$ Now we want to complete $x^2 - x$ into a perfect square. To do that, we should add $\left(\dfrac{{-1}}{2}\right)^2={\dfrac{1}{4}}$ to it: $x^2{-}x+{\dfrac{1}{4}}=\left(x -\dfrac{1}{2}\right)^2$ We add ${\dfrac{1}{4}}$ inside the parentheses, and subtract ${1}\cdot{\dfrac{1}{4}}$ outside them, to keep the expression equivalent. $\begin{aligned} &\phantom{=} \left(x^2 - x\right) -6 \\\\ &=\left(x^2 - x+{\dfrac{1}{4}}\right) -6 -{1}\cdot{\dfrac{1}{4}} \\\\ &= \left(x -\dfrac{1}{2}\right)^2 -6 -\dfrac{1}{4} \\\\ &= \left(x -\dfrac{1}{2}\right)^2 -\dfrac{25}{4} \end{aligned}$ In conclusion, the function after completing the square is written as: $g(x)= \left(x -\dfrac{1}{2}\right)^2 -\dfrac{25}{4}$